Related
I essentially have two columns (vectors) with speed and accel in a data.frame as such:
speed acceleration
1 3.2694444 2.6539535522
2 3.3388889 2.5096979141
3 3.3888889 2.2722134590
4 3.4388889 1.9815256596
5 3.5000000 1.6777544022
6 3.5555556 1.3933215141
7 3.6055556 1.1439051628
8 3.6527778 0.9334115982
9 3.6722222 0.7561602592
I need to find for each value speed on the x axis (speed), what is the top 10% max values from the y axis (acceleration). This also needs to be in a specific interval. For example speed 3.2-3.4, 3.4-3.6, and so on. Can you please show me how a for loop would look like in this situation?
As #alistaire already pointed out, you have provided a very limited amount of data. So we first have to simulate I a bit more data based on which we can test our code.
set.seed(1)
# your data
speed <- c(3.2694444, 3.3388889, 3.3388889, 3.4388889, 3.5,
3.5555556, 3.6055556, 3.6527778, 3.6722222)
acceleration <- c(2.6539535522, 2.5096979141, 2.2722134590,
1.9815256596, 1.6777544022, 1.3933215141,
1.1439051628, 0.9334115982, 0.7561602592)
df <- data.frame(speed, acceleration)
# expand data.frame and add a little bit of noise to all values
# to make them 'unique'
df <- as.data.frame(do.call(
rbind,
replicate(15L, apply(df, 2, \(x) (x + runif(length(x), -1e-1, 1e-1) )),
simplify = FALSE)
))
The function create_intervals, as the name suggests, creates user-defined intervals. The rest of the code does the 'heavy lifting' and stores the desired result in out.
If you would like to have intervals of speed with equal widths, simply specify the number of groups (n_groups) you would like to have and leave the rest of the arguments (i.e. lwr, upr, and interval_span) unspecified.
# Cut speed into user-defined intervals
create_intervals <- \(n_groups = NULL, lwr = NULL, upr = NULL, interval_span = NULL) {
if (!is.null(lwr) & !is.null(upr) & !is.null(interval_span) & is.null(n_groups)) {
speed_low <- subset(df, speed < lwr, select = speed)
first_interval <- with(speed_low, c(min(speed), lwr))
middle_intervals <- seq(lwr + interval_span, upr - interval_span, interval_span)
speed_upp <- subset(df, speed > upr, select = speed)
last_interval <- with(speed_upp, c(upr, max(speed)))
intervals <- c(first_interval, middle_intervals, last_interval)
} else {
step <- with(df, c(max(speed) - min(speed))/n_groups)
intervals <- array(0L, dim = n_groups)
for(i in seq_len(n_groups)) {
intervals[i] <- min(df$speed) + i * step
}
}
return(intervals)
}
# three intervals with equal width
my_intervals <- create_intervals(n_groups = 3L)
# Compute values of speed when acceleration is greater then
# or equal to the 90th percentile
out <- lapply(1:(length(my_intervals)-1L), \(i) {
x <- subset(df, speed >= my_intervals[i] & speed <= my_intervals[i+1L])
x[x$acceleration >= quantile(x$acceleration, 0.9), ]
})
# function to round values to two decimal places
r <- \(x) format(round(x, 2), nsmall = 2L)
# assign names to each element of out
for(i in seq_along(out)) {
names(out)[i] <- paste0(r(my_intervals[i]), '-', r(my_intervals[i+1L]))
}
Output 1
> out
$`3.38-3.57`
speed acceleration
11 3.394378 2.583636
21 3.383631 2.267659
57 3.434123 2.300234
83 3.394886 2.580924
101 3.395459 2.460971
$`3.57-3.76`
speed acceleration
6 3.635234 1.447290
41 3.572868 1.618293
51 3.615017 1.420020
95 3.575412 1.763215
We could also compute the desired values of speed based on intervals that make more 'sense' than just equally spaced speed intervals, e.g. [min(speed), 3.3), [3.3, 3.45), [3.45, 3.6), and [3.6, max(speed)).
This can be accomplished by leaving n_groups unspecified and instead specify lwr, upr, and an interval_span that makes sense. For instance, it makes sense to have a interval span of 0.15 when the lower limit is 3.3 and the upper limit is 3.6.
# custom boundaries based on a lower limit and upper limit
my_intervals <- create_intervals(lwr = 3.3, upr = 3.6, interval_span = 0.15)
Output 2
> out
$`3.18-3.30`
speed acceleration
37 3.238781 2.696456
82 3.258691 2.722076
$`3.30-3.45`
speed acceleration
11 3.394378 2.583636
19 3.328292 2.711825
73 3.315306 2.644580
83 3.394886 2.580924
$`3.45-3.60`
speed acceleration
4 3.520530 2.018930
40 3.517329 2.032943
58 3.485247 2.079893
67 3.458031 2.078545
$`3.60-3.76`
speed acceleration
6 3.635234 1.447290
34 3.688131 1.218969
51 3.615017 1.420020
78 3.628465 1.348873
Note: use function(x) instead of \(x) if you use a version of R <4.1.0
When using the matchit-function for full matching, the results differ by the order of the input dataframe. That is, if the order of the data is changed, results change, too. This is surprising, because in my understanding, the optimal full algorithm should yield only one single best solution.
Am I missing something or is this an error?
Similar differences occur with the optimal algorithm.
Below you find a reproducible example. Subclasses should be identical for the two data sets, which they are not.
Thank you for your help!
# create data
nr <- c(1:100)
x1 <- rnorm(100, mean=50, sd=20)
x2 <- c(rep("a", 20),rep("b", 60), rep("c", 20))
x3 <- rnorm(100, mean=230, sd=2)
outcome <- rnorm(100, mean=500, sd=20)
group <- c(rep(0, 50),rep(1, 50))
df <- data.frame(x1=x1, x2=x2, outcome=outcome, group=group, row.names=nr, nr=nr)
df_neworder <- df[order(outcome),] # re-order data.frame
# perform matching
model_oldorder <- matchit(group~x1, data=df, method="full", distance ="logit")
model_neworder <- matchit(group~x1, data=df_neworder, method="full", distance ="logit")
# store matching results
matcheddata_oldorder <- match.data(model_oldorder, distance="pscore")
matcheddata_neworder <- match.data(model_neworder, distance="pscore")
# Results based on original data.frame
head(matcheddata_oldorder[order(nr),], 10)
x1 x2 outcome group nr pscore weights subclass
1 69.773776 a 489.1769 0 1 0.5409943 1.0 27
2 63.949637 a 529.2733 0 2 0.5283582 1.0 32
3 52.217666 a 526.7928 0 3 0.5028106 0.5 17
4 48.936397 a 492.9255 0 4 0.4956569 1.0 9
5 36.501507 a 512.9301 0 5 0.4685876 1.0 16
# Results based on re-ordered data.frame
head(matcheddata_neworder[order(matcheddata_neworder$nr),], 10)
x1 x2 outcome group nr pscore weights subclass
1 69.773776 a 489.1769 0 1 0.5409943 1.0 25
2 63.949637 a 529.2733 0 2 0.5283582 1.0 31
3 52.217666 a 526.7928 0 3 0.5028106 0.5 15
4 48.936397 a 492.9255 0 4 0.4956569 1.0 7
5 36.501507 a 512.9301 0 5 0.4685876 2.0 14
Apparently, the assignment of objects to subclasses differs. In my understanding, this should not be the case.
The developers of the optmatch package (which the matchit function calls) provided useful help:
I think what we're seeing here is the result of the tolerance argument
that fullmatch has. The matching algorithm requires integer distances,
so we have to scale then truncate floating point distances. For a
given set of integer distances, there may be multiple matchings that
achieve the minimum, so the solver is free to pick among these
non-unique solutions.
Developing your example a little more:
> library(optmatch)
> nr <- c(1:100) x1 <- rnorm(100, mean=50, sd=20)
> outcome <- rnorm(100, mean=500, sd=20) group <- c(rep(0, 50),rep(1, 50))
> df_oldorder <- data.frame(x1=x1, outcome=outcome, group=group, row.names=nr, nr=nr) > df_neworder <- df_oldorder[order(outcome),] # > re-order data.frame
> glm_oldorder <- match_on(glm(group~x1, > data=df_oldorder), data = df_oldorder)
> glm_neworder <- > match_on(glm(group~x1, data=df_neworder), data = df_neworder)
> fm_old <- fullmatch(glm_oldorder, data=df_oldorder)
> fm_new <- fullmatch(glm_neworder, data=df_neworder)
> mean(sapply(matched.distances(fm_old, glm_oldorder), mean))
> ## 0.06216174
> mean(sapply(matched.distances(fm_new, glm_neworder), mean))
> ## 0.062058 mean(sapply(matched.distances(fm_old, glm_oldorder), mean)) -
> mean(sapply(matched.distances(fm_new, glm_neworder), mean))
> ## 0.00010373
which we can see is smaller than the default tolerance of 0.001. You can always decrease the tolerance level, which may
require increased run time, in order to get closer to the true
floating put minimum. We found 0.001 seemed to work well in practice,
but there is nothing special about this value.
I am trying to apply the Simpson's Diversity Index across a number of different datasets with a variable number of species ('nuse') captured. As such I am trying to construct code which can cope with this automatically without needing to manually construct a formula each time I do it. Example dataset for a manual formula is below:
diverse <- data.frame(nuse1=c(0,20,40,20), nuse2=c(5,5,3,20), nuse3=c(0,2,8,20), nuse4=c(5,8,2,20), total=c(10,35,53,80))
simp <- function(x) {
total <- x[,"total"]
nuse1 <- x[,"nuse1"]
nuse2 <- x[,"nuse2"]
nuse3 <- x[,"nuse3"]
nuse4 <- x[,"nuse4"]
div <- round(((1-(((nuse1*(nuse1 - 1)) + (nuse2*(nuse2 - 1)) + (nuse3*(nuse3 - 1)) + (nuse4*(nuse4 - 1)))/(total*(total - 1))))),digits=4)
return(div)
}
diverse$Simpson <- simp(diverse)
diverse
As you can see this works fine. However, how would I be able to create a function which could automatically adjust to, for example, 9 species (so up to nuse9)?
I have experimented with the paste function + as.formula as indicated here Formula with dynamic number of variables; however it is the expand form of (nuse1 * (nuse1 - 1)) that I'm struggling with. Does anyone have any suggestions please? Thanks.
How about something like:
diverse <- data.frame(nuse1=c(0,20,40,20), nuse2=c(5,5,3,20), nuse3=c(0,2,8,20), nuse4=c(5,8,2,20), total=c(10,35,53,80))
simp <- function(x, species) {
spcs <- grep(species, colnames(x)) # which column names have "nuse"
total <- rowSums(x[,spcs]) # sum by row
div <- round(1 - rowSums(apply(x[,spcs], 2, function(s) s*(s-1))) / (total*(total - 1)), digits = 4)
return(div)
}
diverse$Simpson2 <- simp(diverse, species = "nuse")
diverse
# nuse1 nuse2 nuse3 nuse4 total Simpson2
# 1 0 5 0 5 10 0.5556
# 2 20 5 2 8 35 0.6151
# 3 40 3 8 2 53 0.4107
# 4 20 20 20 20 80 0.7595
All it does is find out which columns start with "nuse" or any other species you have in your dataset. It constructs the "total" value within the function and does not require a total column in the dataset.
I have to calculate cosine similarity (patient similarity metric) in R between 48k patients data with some predictive variables. Here is the equation: PSM(P1,P2) = P1.P2/ ||P1|| ||P2||
where P1 and P2 are the predictor vectors corresponding to two different patients, where for example P1 index patient and P2 will be compared with index (P1) and finally pairwise patient similarity metric PSM(P1,P2) will be calculated.
This process will go on for all 48k patients.
I have added sample data-set for 300 patients in a .csv file. Please find the sample data-set here.https://1drv.ms/u/s!AhoddsPPvdj3hVTSbosv2KcPIx5a
First things first: You can find more rigorous treatments of cosine similarity at either of these posts:
Find cosine similarity between two arrays
Creating co-occurrence matrix
Now, you clearly have a mixture of data types in your input, at least
decimal
integer
categorical
I suspect that some of the integer values are Booleans or additional categoricals. Generally, it will be up to you to transform these into continuous numerical vectors if you want to use them as input into the similarity calculation. For example, what's the distance between admission types ELECTIVE and EMERGENCY? Is it a nominal or ordinal variable? I will only be modelling the columns that I trust to be numerical dependent variables.
Also, what have you done to ensure that some of your columns don't correlate with others? Using just a little awareness of data science and biomedical terminology, it seems likely that the following are all correlated:
diasbp_max, diasbp_min, meanbp_max, meanbp_min, sysbp_max and sysbp_min
I suggest going to a print shop and ordering a poster-size printout of psm_pairs.pdf. :-) Your eyes are better at detecting meaningful (but non-linear) dependencies between variable. Including multiple measurements of the same fundamental phenomenon may over-weight that phenomenon in your similarity calculation. Don't forget that you can derive variables like
diasbp_rage <- diasbp_max - diasbp_min
Now, I'm not especially good at linear algebra, so I'm importing a cosine similarity function form the lsa text analysis package. I'd love to see you write out the formula in your question as an R function. I would write it to compare one row to another, and use two nested apply loops to get all comparisons. Hopefully we'll get the same results!
After calculating the similarity, I try to find two different patients with the most dissimilar encounters.
Since you're working with a number of rows that's relatively large, you'll want to compare various algorithmic methodologies for efficiency. In addition, you could use SparkR/some other Hadoop solution on a cluster, or the parallel package on a single computer with multiple cores and lots of RAM. I have no idea whether the solution I provided is thread-safe.
Come to think of it, the transposition alone (as I implemented it) is likely to be computationally costly for a set of 1 million patient-encounters. Overall, (If I remember my computational complexity correctly) as the number of rows in your input increases, the performance could degrade exponentially.
library(lsa)
library(reshape2)
psm_sample <- read.csv("psm_sample.csv")
row.names(psm_sample) <-
make.names(paste0("patid.", as.character(psm_sample$subject_id)), unique = TRUE)
temp <- sapply(psm_sample, class)
temp <- cbind.data.frame(names(temp), as.character(temp))
names(temp) <- c("variable", "possible.type")
numeric.cols <- (temp$possible.type %in% c("factor", "integer") &
(!(grepl(
pattern = "_id$", x = temp$variable
))) &
(!(
grepl(pattern = "_code$", x = temp$variable)
)) &
(!(
grepl(pattern = "_type$", x = temp$variable)
))) | temp$possible.type == "numeric"
psm_numerics <- psm_sample[, numeric.cols]
row.names(psm_numerics) <- row.names(psm_sample)
psm_numerics$gender <- as.integer(psm_numerics$gender)
psm_scaled <- scale(psm_numerics)
pair.these.up <- psm_scaled
# checking for independence of variables
# if the following PDF pair plot is too big for your computer to open,
# try pair-plotting some random subset of columns
# keep.frac <- 0.5
# keep.flag <- runif(ncol(psm_scaled)) < keep.frac
# pair.these.up <- psm_scaled[, keep.flag]
# pdf device sizes are in inches
dev <-
pdf(
file = "psm_pairs.pdf",
width = 50,
height = 50,
paper = "special"
)
pairs(pair.these.up)
dev.off()
#transpose the dataframe to get the
#similarity between patients
cs <- lsa::cosine(t(psm_scaled))
# this is super inefficnet, because cs contains
# two identical triangular matrices
cs.melt <- melt(cs)
cs.melt <- as.data.frame(cs.melt)
names(cs.melt) <- c("enc.A", "enc.B", "similarity")
extract.pat <- function(enc.col) {
my.patients <-
sapply(enc.col, function(one.pat) {
temp <- (strsplit(as.character(one.pat), ".", fixed = TRUE))
return(temp[[1]][[2]])
})
return(my.patients)
}
cs.melt$pat.A <- extract.pat(cs.melt$enc.A)
cs.melt$pat.B <- extract.pat(cs.melt$enc.B)
same.pat <- cs.melt[cs.melt$pat.A == cs.melt$pat.B ,]
different.pat <- cs.melt[cs.melt$pat.A != cs.melt$pat.B ,]
most.dissimilar <-
different.pat[which.min(different.pat$similarity),]
dissimilar.pat.frame <- rbind(psm_numerics[rownames(psm_numerics) ==
as.character(most.dissimilar$enc.A) ,],
psm_numerics[rownames(psm_numerics) ==
as.character(most.dissimilar$enc.B) ,])
print(t(dissimilar.pat.frame))
which gives
patid.68.49 patid.9
gender 1.00000 2.00000
age 41.85000 41.79000
sysbp_min 72.00000 106.00000
sysbp_max 95.00000 217.00000
diasbp_min 42.00000 53.00000
diasbp_max 61.00000 107.00000
meanbp_min 52.00000 67.00000
meanbp_max 72.00000 132.00000
resprate_min 20.00000 14.00000
resprate_max 35.00000 19.00000
tempc_min 36.00000 35.50000
tempc_max 37.55555 37.88889
spo2_min 90.00000 95.00000
spo2_max 100.00000 100.00000
bicarbonate_min 22.00000 26.00000
bicarbonate_max 22.00000 30.00000
creatinine_min 2.50000 1.20000
creatinine_max 2.50000 1.40000
glucose_min 82.00000 129.00000
glucose_max 82.00000 178.00000
hematocrit_min 28.10000 37.40000
hematocrit_max 28.10000 45.20000
potassium_min 5.50000 2.80000
potassium_max 5.50000 3.00000
sodium_min 138.00000 136.00000
sodium_max 138.00000 140.00000
bun_min 28.00000 16.00000
bun_max 28.00000 17.00000
wbc_min 2.50000 7.50000
wbc_max 2.50000 13.70000
mingcs 15.00000 15.00000
gcsmotor 6.00000 5.00000
gcsverbal 5.00000 0.00000
gcseyes 4.00000 1.00000
endotrachflag 0.00000 1.00000
urineoutput 1674.00000 887.00000
vasopressor 0.00000 0.00000
vent 0.00000 1.00000
los_hospital 19.09310 4.88130
los_icu 3.53680 5.32310
sofa 3.00000 5.00000
saps 17.00000 18.00000
posthospmort30day 1.00000 0.00000
Usually I wouldn't add a second answer, but that might be the best solution here. Don't worry about voting on it.
Here's the same algorithm as in my first answer, applied to the iris data set. Each row contains four spatial measurements of the flowers form three different varieties of iris plants.
Below that you will find the iris analysis, written out as nested loops so you can see the equivalence. But that's not recommended for production with large data sets.
Please familiarize yourself with starting data and all of the intermediate dataframes:
The input iris data
psm_scaled (the spatial measurements, scaled to mean=0, SD=1)
cs (the matrix of pairwise similarities)
cs.melt (the pairwise similarities in long format)
At the end I have aggregated the mean similarities for all comparisons between one variety and another. You will see that comparisons between individuals of the same variety have mean similarities approaching 1, and comparisons between individuals of the same variety have mean similarities approaching negative 1.
library(lsa)
library(reshape2)
temp <- iris[, 1:4]
iris.names <- paste0(iris$Species, '.', rownames(iris))
psm_scaled <- scale(temp)
rownames(psm_scaled) <- iris.names
cs <- lsa::cosine(t(psm_scaled))
# this is super inefficient, because cs contains
# two identical triangular matrices
cs.melt <- melt(cs)
cs.melt <- as.data.frame(cs.melt)
names(cs.melt) <- c("enc.A", "enc.B", "similarity")
names(cs.melt) <- c("flower.A", "flower.B", "similarity")
class.A <-
strsplit(as.character(cs.melt$flower.A), '.', fixed = TRUE)
cs.melt$class.A <- sapply(class.A, function(one.split) {
return(one.split[1])
})
class.B <-
strsplit(as.character(cs.melt$flower.B), '.', fixed = TRUE)
cs.melt$class.B <- sapply(class.B, function(one.split) {
return(one.split[1])
})
cs.melt$comparison <-
paste0(cs.melt$class.A , '_vs_', cs.melt$class.B)
cs.agg <-
aggregate(cs.melt$similarity, by = list(cs.melt$comparison), mean)
print(cs.agg[order(cs.agg$x),])
which gives
# Group.1 x
# 3 setosa_vs_virginica -0.7945321
# 7 virginica_vs_setosa -0.7945321
# 2 setosa_vs_versicolor -0.4868352
# 4 versicolor_vs_setosa -0.4868352
# 6 versicolor_vs_virginica 0.3774612
# 8 virginica_vs_versicolor 0.3774612
# 5 versicolor_vs_versicolor 0.4134413
# 9 virginica_vs_virginica 0.7622797
# 1 setosa_vs_setosa 0.8698189
If you’re still not comfortable with performing lsa::cosine() on a scaled, numerical dataframe, we can certainly do explicit pairwise calculations.
The formula you gave for PSM, or cosine similarity of patients, is expressed in two formats at Wikipedia
Remembering that vectors A and B represent the ordered list of attributes for PatientA and PatientB, the PSM is the dot product of A and B, divided by (the scalar product of [the magnitude of A] and [the magnitude of B])
The terse way of saying that in R is
cosine.sim <- function(A, B) { A %*% B / sqrt(A %*% A * B %*% B) }
But we can rewrite that to look more similar to your post as
cosine.sim <- function(A, B) { A %*% B / (sqrt(A %*% A) * sqrt(B %*% B)) }
I guess you could even re-write that (the calculations of similarity between a single pair of individuals) as a bunch of nested loops, but in the case of a manageable amount of data, please don’t. R is highly optimized for operations on vectors and matrices. If you’re new to R, don’t second guess it. By the way, what happened to your millions of rows? This will certainly be less stressful now that your down to tens of thousands.
Anyway, let’s say that each individual only has two elements.
individual.1 <- c(1, 0)
individual.2 <- c(1, 1)
So you can think of individual.1 as a line that passes between the origin (0,0) and (0, 1) and individual.2 as a line that passes between the origin and (1, 1).
some.data <- rbind.data.frame(individual.1, individual.2)
names(some.data) <- c('element.i', 'element.j')
rownames(some.data) <- c('individual.1', 'individual.2')
plot(some.data, xlim = c(-0.5, 2), ylim = c(-0.5, 2))
text(
some.data,
rownames(some.data),
xlim = c(-0.5, 2),
ylim = c(-0.5, 2),
adj = c(0, 0)
)
segments(0, 0, x1 = some.data[1, 1], y1 = some.data[1, 2])
segments(0, 0, x1 = some.data[2, 1], y1 = some.data[2, 2])
So what’s the angle between vector individual.1 and vector individual.2? You guessed it, 0.785 radians, or 45 degrees.
cosine.sim <- function(A, B) { A %*% B / (sqrt(A %*% A) * sqrt(B %*% B)) }
cos.sim.result <- cosine.sim(individual.1, individual.2)
angle.radians <- acos(cos.sim.result)
angle.degrees <- angle.radians * 180 / pi
print(angle.degrees)
# [,1]
# [1,] 45
Now we can use the cosine.sim function I previously defined, in two nested loops, to explicitly calculate the pairwise similarities between each of the iris flowers. Remember, psm_scaled has already been defined as the scaled numerical values from the iris dataset.
cs.melt <- lapply(rownames(psm_scaled), function(name.A) {
inner.loop.result <-
lapply(rownames(psm_scaled), function(name.B) {
individual.A <- psm_scaled[rownames(psm_scaled) == name.A, ]
individual.B <- psm_scaled[rownames(psm_scaled) == name.B, ]
similarity <- cosine.sim(individual.A, individual.B)
return(list(name.A, name.B, similarity))
})
inner.loop.result <-
do.call(rbind.data.frame, inner.loop.result)
names(inner.loop.result) <-
c('flower.A', 'flower.B', 'similarity')
return(inner.loop.result)
})
cs.melt <- do.call(rbind.data.frame, cs.melt)
Now we repeat the calculation of cs.melt$class.A, cs.melt$class.B, and cs.melt$comparison as above, and calculate cs.agg.from.loops as the mean similarity between the various types of comparisons:
cs.agg.from.loops <-
aggregate(cs.agg.from.loops$similarity, by = list(cs.agg.from.loops $comparison), mean)
print(cs.agg.from.loops[order(cs.agg.from.loops$x),])
# Group.1 x
# 3 setosa_vs_virginica -0.7945321
# 7 virginica_vs_setosa -0.7945321
# 2 setosa_vs_versicolor -0.4868352
# 4 versicolor_vs_setosa -0.4868352
# 6 versicolor_vs_virginica 0.3774612
# 8 virginica_vs_versicolor 0.3774612
# 5 versicolor_vs_versicolor 0.4134413
# 9 virginica_vs_virginica 0.7622797
# 1 setosa_vs_setosa 0.8698189
Which, I believe is identical to the result we got with lsa::cosine.
So what I'm trying to say is... why wouldn't you use lsa::cosine?
Maybe you should be more concerned with
selection of variables, including removal of highly correlated variables
scaling/normalizing/standardizing the data
performance with a large input data set
identifying known similars and dissimilars for quality control
as previously addressed
I am trying to compute in R. I have the following values.
nb <- 100
tb <- 25
ns <- 90
ts <- 15
A0 <- 1
S_norm <- 0.4
R <-tb/ts
y_meas <- (ns-nb/R)/A0
sigma_meas = sqrt(ns+(nb+1)/R^2)/A0
I am very confused on how I can integrate L(psi), say from -10 to 10. Because I am integrating with respect to log A.
You can substitute for logA and for a fixed value of psi you can integrate as follows:
psi <- 5
integrate(function(x) exp(-0.5*(((x/A0)/S_norm)^2 + ((psi-y_meas*A0/exp(x))/sigma_meas)^2)),
-10, 10)
# 0.1775989 with absolute error < 6.6e-05
On top of an excellent answer by #SandipanDey, if you could extend limits to -Infinity...+Infinity, there is a better way to integrate functions with e-x2 kernel: Gauss-Hermite quadrature, and there is an R package for that.
Simple example:
library(gaussquad)
n.quad <- 128 # integration order
# get the particular (weights,abscissas) as data frame
# with 2 observables and n.quad observations
rule <- ghermite.h.quadrature.rules(n.quad, mu = 0.0)[[n.quad]]
# test function - integrate 1 over exp(-x^2) from -Inf to Inf
# should get sqrt(pi) as an answer
f <- function(x) {
1.0
}
q <- ghermite.h.quadrature(f, rule)
print(q - sqrt(pi))